# If $X$ is maximal ideal then it consists of non-invertible elements?

I'm reading through a paper where I came across the following theorem

Let $A$ be a commutative complex Banach algebra with unit element $e$.

Theorem: A subspace $X\subset A$ of codimension $1$ is a maximal ideal in $A$ if and only if it consists of non-invertible elements.

Proof:

Clearly any maximal ideal satisfies the above condition, so it is sufficient to show that if $\text{codim } X=1$, and if $X$ consists of non-invertible elements, then $X$ is a maximal ideal in $A$.

I understand the rest of the proof, so I will omit the rest. What I feel unsure about in the extract of the proof stated above, is the following:

Why does any maximal ideal clearly satisfy the conditions (ie it consists of non-invertible elements)? My reasoning and trail of thought it something along the following lines:

Non-invertible elements lie in some proper ideal and all proper ideals are contained in maximal ideals. Hence the maximal ideal consists out of non-invertible elements. Is this the correct reasoning? Can anyone perhaps help show me a more mathematically rigorous way of "proving" it?

A maximal ideal is by definition a proper ideal (which happens to not be strictly contained in any other proper ideal). If $I$ is an ideal and an element $x\in I$ is invertible, then $x^{-1}x=e\in I$ so $a=a\cdot e\in I$ for all $a\in A$, so $I=A$. So no proper ideal (and in particular, no maximal ideal) can contain an invertible element.
Since I can't post comments on Mathoverflow, I answer the other part of your question https://mathoverflow.net/questions/233931/showing-that-x-kerg-is-the-maximal-ideal-of-a-from-the-fact-that-it-is here. If $X$ has codimension 1, A is a unital Banach algebra and $X$ consists of non-invertible elements, then $A=X\oplus\mathbb{C}e$. So as you stated, $X$ is the kernel $ker(g)$ of some linear functional. But moreover it's the kernel of a unital algebra homomorphism (Choose $g(e)=1, \left.g\right|_X=0$). But for a unital complex Banach algebra A, there's a 1:1 correspondence between
$\text{{maximal ideals of A}}\,=\,\mathcal{M}(A)\,\overset{1:1}\longleftrightarrow\,\Omega(A)=\text{{continous unital algebra homomorphisms}}$<